Arthur Mattuck, Introduction to Analysis, Problem 5-7

This question is from Arthur Mattuck's "Introduction to Analysis", chapter 5, problem 5-7.

1. The problem statement, all variables and given/known data
Define a sequence recursively by [tex]a_{n+1}=\sqrt{2a_{n}}[/tex], [tex]a_{0}>0[/tex].

(a) Prove that for any choice of [tex]a_{0}>0[/tex], the sequence is monotone and bounded.

2. Relevant equations

None

3. The attempt at a solution

I've rewritten the sequence as

[tex]a_n = \sqrt{2{\sqrt{2\sqrt{2\sqrt{2...\sqrt{2a_0}}}}}[/tex],

and taken the ratio of [tex]a_{n+1}[/tex] and [tex]a_{n}[/tex], which leads me to the expression

[tex]\frac{a_{n+1}}{a_n} = 2^{(1/2)^{n+1}}\sqrt{a_0}[/tex].

From computing the sequence for a few initial values of [tex]a_0[/tex], I've been able to determine that the sequence is decreasing if [tex]a_0>2[/tex] and increasing if [tex]0<a_0<2[/tex]. However, I'm not sure how to show this from this ratio, i.e. how does

[tex]2^{(1/2)^{n+1}}\sqrt{a_0}<1[/tex]

which indicates that the sequence is decreasing, turn into [tex]a_0>2[/tex]?

The rest of the problem comes easy enough. I'm able to show that it is bounded and that the limit is 2 (in part b of the question).